Given that,
mean(x)=10.2
standard deviation , sigma1 =2.7
number(n1)=30
y(mean)=7.8
standard deviation, sigma2 =2.5
number(n2)=30
null, Ho: u1 = u2
alternate, H1: μ1 != u2
level of significance, alpha = 0.05
from standard normal table, two tailed z alpha/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=10.2-7.8/sqrt((7.29/30)+(6.25/30))
zo =3.572
| zo | =3.572
critical value
the value of |z alpha| at los 0.05% is 1.96
we got |zo | =3.572 & | z alpha | =1.96
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.572 ) =
0.00035
hence value of p0.05 > 0.00035,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: μ1 != u2
c.
test statistic: 3.572
critical value: -1.96 , 1.96
b.
p-value: 0.00035
decision: reject Ho
d.
TRADITIONAL METHOD
given that,
mean(x)=10.2
standard deviation , σ1 =2.7
population size(n1)=30
y(mean)=7.8
standard deviation, σ2 =2.5
population size(n2)=30
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((7.29/30)+(6.25/30))
= 0.6718
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.6718
= 1.3168
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (10.2-7.8) ± 1.3168 ]
= [1.0832 , 3.7168]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=10.2
standard deviation , σ1 =2.7
number(n1)=30
y(mean)=7.8
standard deviation, σ2 =2.5
number(n2)=30
CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 10.2-7.8) ±Z a/2 * Sqrt( 7.29/30+6.25/30)]
= [ (2.4) ± Z a/2 * Sqrt( 0.4513) ]
= [ (2.4) ± 1.96 * Sqrt( 0.4513) ]
= [1.0832 , 3.7168]
-----------------------------------------------------------------------------------------------
e.
interpretations:
1. we are 95% sure that the interval [1.0832 , 3.7168] contains the
difference between
true population mean U1 - U2
2. If a large number of samples are collected, and a confidence
interval is created
for each sample, 95% of these intervals will contains the
difference between
true population mean U1 - U2
3. Since this Cl does contain a zero we can conclude at 0.05 true
mean
difference is zero.
conclusion:
we have enough evidence to support the claim that difference of
means males and females of spatial assesement.
f.
Given that,
mean(x)=10.9
standard deviation , sigma1 =3
number(n1)=30
y(mean)=15.1
standard deviation, sigma2 =3.4
number(n2)=30
null, Ho: u1 = u2
alternate, H1: μ1 != u2
level of significance, alpha = 0.05
from standard normal table, two tailed z alpha/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=10.9-15.1/sqrt((9/30)+(11.56/30))
zo =-5.073
| zo | =5.073
critical value
the value of |z alpha| at los 0.05% is 1.96
we got |zo | =5.073 & | z alpha | =1.96
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -5.073 )
= 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
i.
null, Ho: u1 = u2
alternate, H1: μ1 != u2
iii.
test statistic: -5.073
critical value: -1.96 , 1.96
decision: reject Ho
ii.
p-value: 0
iv.
TRADITIONAL METHOD
given that,
mean(x)=10.9
standard deviation , σ1 =3
population size(n1)=30
y(mean)=15.1
standard deviation, σ2 =3.4
population size(n2)=30
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((9/30)+(11.56/30))
= 0.8278
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.8278
= 1.6226
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (10.9-15.1) ± 1.6226 ]
= [-5.8226 , -2.5774]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=10.9
standard deviation , σ1 =3
number(n1)=30
y(mean)=15.1
standard deviation, σ2 =3.4
number(n2)=30
CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 10.9-15.1) ±Z a/2 * Sqrt( 9/30+11.56/30)]
= [ (-4.2) ± Z a/2 * Sqrt( 0.6853) ]
= [ (-4.2) ± 1.96 * Sqrt( 0.6853) ]
= [-5.8226 , -2.5774]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-5.8226 , -2.5774] contains
the difference between
true population mean U1 - U2
2. If a large number of samples are collected, and a confidence
interval is created
for each sample, 95% of these intervals will contains the
difference between
true population mean U1 - U2
3. Since this Cl does contain a zero we can conclude at 0.05 true
mean
difference is zero
v.
conclusion:
we have enough evidence to support the claim that difference of
means of males and females of verbal assesement
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